Counterexample for a polar decomposition in von Neumann and $C^\ast$ algebras

Question

For a von Neumann algebra, we have that partial isometry and positive operator of an operator in its polar decomposition belongs to the algebra, but in a $C^\ast$ algebra this may not be true.

Can anyone help me find a counterexample?

Answer 1

Consider $C^*$-algebra $C([-1,1])$ and its non invertible element $f(x)=x$, then its partial isometry, which is $\mathrm{sign}(x)$, does not belong to $C([-1,1])$.

For a complex counterexample, note that the same example $f(z)=z$ works in $C(\mathbb{D},\mathbb{C})$, where $\mathbb{D}$ is the closed unit ball in $\mathbb{C}$. Indeed, if $f=u|f|$ is a polar decomposition, then $$ u(z)=\frac{z}{|z|}\quad\forall z\neq 0 $$ does not have a limit at $0$. So whatever partial isometry $u$ you take, it will never be continuous.